我正在用jQuery改变CSS，我希望删除我根据输入值添加的样式:
if(color != '000000') $("body").css("background-color", color); else // remove style ?
我该怎么做呢? 请注意，每当使用颜色选择器选择颜色时，上面的行就会运行。当鼠标移动到一个色轮上时)。
第二注意:我不能这样做与css("background-color"， "none")，因为它将删除默认样式从css文件。 我只是想删除jQuery添加的背景色内联样式。
当前回答
这比其他一些解决方案更复杂，但可能在场景中提供更大的灵活性:
1)做一个类定义来隔离(封装)你想要有选择地应用/删除的样式。它可以是空的(对于这种情况，可能应该是空的):
.myColor {}
2)使用以下代码，基于gilly3的回答http://jsfiddle.net/kdp5V/167/:
function changeCSSRule(styleSelector,property,value) {
for (var ssIdx = 0; ssIdx < document.styleSheets.length; ssIdx++) {
var ss = document.styleSheets[ssIdx];
var rules = ss.cssRules || ss.rules;
if(rules){
for (var ruleIdx = 0; ruleIdx < rules.length; ruleIdx++) {
var rule = rules[ruleIdx];
if (rule.selectorText == styleSelector) {
if(typeof value == 'undefined' || !value){
rule.style.removeProperty(property);
} else {
rule.style.setProperty(property,value);
}
return; // stops at FIRST occurrence of this styleSelector
}
}
}
}
}
使用示例:http://jsfiddle.net/qvkwhtow/
警告:
Not extensively tested. Can't include !important or other directives in the new value. Any such existing directives will be lost through this manipulation. Only changes first found occurrence of a styleSelector. Doesn't add or remove entire styles, but this could be done with something more elaborate. Any invalid/unusable values will be ignored or throw error. In Chrome (at least), non-local (as in cross-site) CSS rules are not exposed through document.styleSheets object, so this won't work on them. One would have to add a local overrides and manipulate that, keeping in mind the "first found" behavior of this code. document.styleSheets is not particularly friendly to manipulation in general, don't expect this to work for aggressive use.
以这种方式隔离样式就是CSS的全部，即使操纵它不是。操纵CSS规则不是jQuery的全部，jQuery操纵DOM元素，并使用CSS选择器来完成它。
其他回答
我讲了用纯JavaScript删除样式属性的方法让你们知道纯JavaScript的方法
var bodyStyle = document.body.style;
if (bodyStyle.removeAttribute)
bodyStyle.removeAttribute('background-color');
else
bodyStyle.removeProperty('background-color');
这比其他一些解决方案更复杂，但可能在场景中提供更大的灵活性:
1)做一个类定义来隔离(封装)你想要有选择地应用/删除的样式。它可以是空的(对于这种情况，可能应该是空的):
.myColor {}
2)使用以下代码，基于gilly3的回答http://jsfiddle.net/kdp5V/167/:
function changeCSSRule(styleSelector,property,value) {
for (var ssIdx = 0; ssIdx < document.styleSheets.length; ssIdx++) {
var ss = document.styleSheets[ssIdx];
var rules = ss.cssRules || ss.rules;
if(rules){
for (var ruleIdx = 0; ruleIdx < rules.length; ruleIdx++) {
var rule = rules[ruleIdx];
if (rule.selectorText == styleSelector) {
if(typeof value == 'undefined' || !value){
rule.style.removeProperty(property);
} else {
rule.style.setProperty(property,value);
}
return; // stops at FIRST occurrence of this styleSelector
}
}
}
}
}
使用示例:http://jsfiddle.net/qvkwhtow/
警告:
Not extensively tested. Can't include !important or other directives in the new value. Any such existing directives will be lost through this manipulation. Only changes first found occurrence of a styleSelector. Doesn't add or remove entire styles, but this could be done with something more elaborate. Any invalid/unusable values will be ignored or throw error. In Chrome (at least), non-local (as in cross-site) CSS rules are not exposed through document.styleSheets object, so this won't work on them. One would have to add a local overrides and manipulate that, keeping in mind the "first found" behavior of this code. document.styleSheets is not particularly friendly to manipulation in general, don't expect this to work for aggressive use.
以这种方式隔离样式就是CSS的全部，即使操纵它不是。操纵CSS规则不是jQuery的全部，jQuery操纵DOM元素，并使用CSS选择器来完成它。
let el = document.querySelector(element)
let styles = el.getAttribute('style')
el.setAttribute('style', styles.replace('width: 100%', ''))
为什么不让你想要删除一个CSS类的样式呢?现在你可以使用:. removeclass()。 这也开启了使用.toggleClass()的可能性:
(如果存在，则删除该类，如果不存在则添加该类。)
在处理布局问题时，添加/删除类也会减少更改/故障排除的混乱(而不是试图找出特定样式消失的原因)。
在我的测试中，接受的答案可以工作，但在DOM上留下一个空的样式属性。没什么大不了的，但这就完全消除了:
removeAttr( 'style' );
这假设您希望删除所有动态样式并返回到样式表样式。
